JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 8)
A man is walking on a straight line. The arithmetic mean
of the reciprocals of the intercepts of this line on the
coordinate axes is $${1 \over 4}$$. Three stones A, B and C are placed at the points
(1, 1), (2, 2) and (4, 4) respectively. Then, which of these stones is / are on the path of the man?
A only
All the three
C only
B only
Explanation
Given, position of A = (1, 1)
Position of B = (2, 2)
Position of C = (4, 4)
_24th_February_Morning_Shift_en_8_1.png)
Let x-intercept be a and y-intercept be b.
Equation of line traced is
$${x \over a} + {y \over b} = 1$$
This is the equation of path, let a point (h, k) lie on this path.
Then, $${h \over a} + {k \over b} = 1$$
Also, AM of reciprocal of a and b = $${1 \over 4}$$
$$\therefore$$ $${{{1 \over a} + {1 \over b}} \over 2} = {1 \over 4}$$
$${1 \over a} + {1 \over b} = {1 \over 2}$$
On comparing Eqs. (i) and (ii), we get (h, k) = (2, 2)
Hence, the required stone is B(2, 2).
Position of B = (2, 2)
Position of C = (4, 4)
Let x-intercept be a and y-intercept be b.
Equation of line traced is
$${x \over a} + {y \over b} = 1$$
This is the equation of path, let a point (h, k) lie on this path.
Then, $${h \over a} + {k \over b} = 1$$
Also, AM of reciprocal of a and b = $${1 \over 4}$$
$$\therefore$$ $${{{1 \over a} + {1 \over b}} \over 2} = {1 \over 4}$$
$${1 \over a} + {1 \over b} = {1 \over 2}$$
On comparing Eqs. (i) and (ii), we get (h, k) = (2, 2)
Hence, the required stone is B(2, 2).
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